Practice Questions

Thank for visiting the EngineerInTrainingExam.com Practice Questions Page for the EIT Exam. I put this these problems together in an effort to help you prepare and get an idea of the types and structure of questions on the exam. For each question, choose the best answer from the four options given. After each question, you will be given a detailed description of how the answer was obtained, so don’t worry if you get it wrong.

Enter you name and email address in the slots given to gain Instant Access

There is no time limit, but it wouldn’t be a bad idea to time yourself just to see where you stand.

Good Luck!

Notice: Undefined variable: mtq_use_timer in /home/thedickm/public_html/engineerintrainingexam.com/wp-content/plugins/mtouch-quiz/show_quiz.php on line 239

Notice: Undefined variable: mtq_use_timer in /home/thedickm/public_html/engineerintrainingexam.com/wp-content/plugins/mtouch-quiz/show_quiz.php on line 360

Please wait while the activity loads. If this activity does not load, try refreshing your browser. Also, this page requires javascript. Please visit using a browser with javascript enabled.

If loading fails, click here to try again

Congratulations!
You scored %%SCORE%% out of %%TOTAL%%.
Your performance has been rated as %%RATING%%
I would be honored to have you as a Subscriber to EngineerInTrainingExam.com. Join the many who receive exclusive content along with the latest tips, guidance, and updates straight to their email. Oh yah, you'll also get my FREE eBook titled "EIT PREPARATION BOOT CAMP; A complete Study Guide to Setting Yourself up for Success on the Engineer In Training Exam "
My strict privacy policy keeps your email address 100% & secure given away, traded, sold, delivered, revealed, publicized, or marketed in any way, shape, or form.

Your answers are highlighted below.

Question 1

Billy Bob's Machine Shop is considering investing \($12,000\) in a machine that produces \($3,500\) annual revenue, in which \(35%\) of that goes towards overhead costs to produce the product. The machine requires \($200\) annual maintenance and has a salvage value of \($900\) after \(8\) years. Disregarding depreciation and assuming \(9%\) interest, determine the present day value of the machine.

A \(12V \) DC source has been running the circuit below for a long time. Given this information, determine the voltage across the Capacitor:

A

$$12\;V$$

B

$$0\;V$$

C

$$6\;V$$

D

$$8.5\;V$$

Question 2 Explanation:

The Voltage across the Capacitor is \(12V\). After a source is connected to a Capacitor for "a long time", the Capacitor becomes fully charged and no more current flows through the terminals at that point, it essentially becomes an open circuit.

Question 3

Determine the solution to the following Differential Equation:
\[
\frac{dy}{dt} = 10-5y \text{ with an initial value of } y(0)=1
\]

Kircoff's Second Law states that for any loop in a circuit, the voltage rises must equal the voltage drops, or in other words, whatever energy a charge starts with in a circuit loop, it must end up losing all that energy by the time it gets to the end so that:
\begin{aligned}
\sum V=0
\end{aligned}
Knowing \(V=IR\), we can annotate the given circuit as such:
\begin{aligned}
\sum {V}=V_1+I(R_{eq1})+I(R_{2})+I(R_{eq3})-V_2=0
\end{aligned}
The equivalent of the resistors in parallel is:
\begin{aligned}
\frac{1}{R_{eq1}}=\frac{1}{10}+\frac{1}{10}=\frac{2}{10}\Rightarrow R_{eq1}=5 \\
\end{aligned}
And
\begin{aligned}
\frac{1}{R_{eq3}}=\frac{1}{10}+\frac{1}{5}=\frac{3}{10}\Rightarrow R_{eq3}=3.33 \\
\end{aligned}
So
\begin{aligned}
\sum V=48+I(5)+I(20)+I(3.33)-24=0
\end{aligned}
Solving for I:
\begin{aligned}
I=\frac{-24}{28.33} \\
\end{aligned}
ANSWER:
\begin{aligned}
I=.85\;amps
\end{aligned}

Question 5

What is the stress intensity factor around an edge crack of length \(2.5\;mm\) in a thick plate of material under an applied stress of \(145\;MPa\)? Assume a geometrical factor of \(1.25\)

Using the Gibbs Phase Rule, determine how many degrees of freedom exist at the triple point for water

A

$$0$$

B

$$2$$

C

$$1$$

D

$$3$$

Question 6 Explanation:

Gibbs Phase rule states that:
\begin{aligned}
P+F=C+2
\end{aligned}
Where:
\begin{aligned}
&\text{P: Number of phases (solid, liquid, or gas)}\\
&\text{C: Number of components}\\
&\text{F: Degrees of Freedom}\\
\end{aligned}
At the triple point of water:
\begin{aligned}
&\text{P = 3 (solid, liquid, and gas)} \\
&\text{C = 1 (water)}
\end{aligned}
Plugging the values in to the equation and rearranging:
\begin{aligned}
3+F=1+2
\end{aligned}
ANSWER:
\begin{aligned}
F=0 \text{ there are zero degrees of freedom at the triple point of water}
\end{aligned}

Question 7

Evaluate the Centroidal Moment of Inertia about an axis parallel to that of the x axis for the object below:

A

$$213.3$$

B

$$621.7$$

C

$$273.1$$

D

$$571$$

Question 7 Explanation:

The First step is to divide the object in to three parts:
$$\\$$
$$\\$$
Determine A and \(\bar{y}\) for each shape:
\begin{matrix}
A_1=1(4.5) & A_2=8(2) & A_3=.5(5.5) \\
\bar{y_1}=.5+8+.5=9 & \bar{y_2}=.5+4=4.5 & \bar{y_2}=.25 \\
A_1\bar{y_1}=40.5 & A_2\bar{y_2}=72 & A_3\bar{y_3}=.69 \\
\end{matrix}\\
Plug in the values to determine} \(\bar{y}_{composite}\):
\begin{aligned}
\bar{y}_{composite}=\frac{\sum{A\bar{y}}}{\sum{A}}=\frac{40.5+72+.69}{4.5+16+2.75}=4.87
\end{aligned}
Determine the Centroidal Moment of Inertia for each shape and the distance from \(\bar{y}_{composite}\) for each shape noting that \( I_{rec}=\frac{bh^3}{12}\) :
\begin{matrix}
I_{c1}=\frac{4.5(1)^3}{12}=.375 & d_1=4.87-9=-4.13 \\
I_{c2}=\frac{2(8)^3}{12}=85.33 & d_2=.37 \\
I_{c3}=\frac{5.5(.5)^3}{12}=.057 & d_3=4.18 \\
\end{matrix}
Utilizing the Parallel Axis Theorem to get the Centroidal Moment of Inertia of the complete object we get:
\begin{aligned}
I=.375+4.5(-4.13)^2+85.33+16(.37)^2+.57+2.75(4.18)^2=213.3
\end{aligned}

Question 8

Find the equation of a circle that passes through each of the given points below:
\begin{aligned}
&A(-5,0) \\
&B(1,0) \\
&C(-2,-3) \\
\end{aligned}

A

$$(x+2)^2+y^2=9$$

B

$$x^2+y^2=7$$

C

$$x^2+(y+2)^2=7$$

D

$$x^2+(y+2)^2=9$$

Question 8 Explanation:

Using the distance formula, we can formulate an equation that represents the distance, R, from each point to the origin, O:
\begin{aligned}
&| AO |=\sqrt{(h+5)^2+(k)^2} \\
&| BO |=\sqrt{(h-1)^2+(k)^2} \\
&| CO |=\sqrt{(h+2)^2+(k+3)^2} \\
\end{aligned}
At this point we have 3 equations and 3 unknowns (h,k,and R)
Knowing that the distance from each point is equal to R, we can set the equations equal to one another to eliminate the variable R for now:
\begin{aligned}
&\sqrt{(h+5)^2+(k)^2}=\sqrt{(h-1)^2+(k)^2} \\
&\sqrt{(h-1)^2+(k)^2}=\sqrt{(h+2)^2+(k+3)^2} \\
\end{aligned}
Squaring both sides and expanding we get:
\begin{aligned}
&(h+5)^2+(k)^2=(h-1)^2+(k)^2\Rightarrow h^2+10h+25+k^2=h^2-2h+1+k^2 \\
&(h-1)^2+(k)^2=(h+2)^2+(k+3)^2\Rightarrow h^2-2h+1+k^2=h^2+4h+4+k^2+6k+9 \\
\end{aligned}
Combining the terms in the first series:
\begin{aligned}
12h=-24 \text{ which gives} h=-2
\end{aligned}
Combining the terms in the second series:
\begin{aligned}
-6h-6k=12
\end{aligned}
Plugging in the value for h:
\begin{aligned}
-6(-2)-6k=12 \text{ which gives} k=0
\end{aligned}
Find R using the values of h and k and any of the original Distance equations you formulated in the first step:
\begin{aligned}
R^2=(-2+5)^2+(0)^2=9
\end{aligned}
Finally, insert all the derived values in to the general equation of a circle:
\begin{aligned}
(x+2)^2+(y-0)^2=9
\end{aligned}
ANSWER:
\begin{aligned}
(x+2)^2+y^2=9
\end{aligned}

Question 9

Cargo is being pulled up a ramp as shown in the figure below. The contents within are fragile so it's crucial the container is pulled up at a constant acceleration of \(3 \frac{m}{s^2}\). Determine the Tension necessary on the backside of the pulley (right side) to ensure the \(100kg\) contents are not disturbed.

Referring to the Ideal Compression Cycle diagrams below, determine which of the general formulas is not correct:

A

$$\eta_t= \frac{W_{cyc}}{Q_{cyc}}$$

B

$$W_{cyc}=Q_{23}+-Q_{41}$$

C

$$W_{cyc}=W_{23}+W_{34}-W_{12}$$

D

$$Q_{cyc}=Q_{23}-Q_{41}$$

Question 10 Explanation:

The efficiency of an Ideal Compression Cycle is \( \eta_t= \frac{W_{cyc}}{Q_{in}} \). If it was the way it is relayed in the option above, the efficiency would be \( \eta_t=\frac{Q_{23}-Q_{41}}{Q_{23}-Q_{41}} \) which would make for a \(100\% \) efficient cycle, which is not possible.

Once you are finished, click the button below. Any items you have not completed will be marked incorrect.
Get Results

There are 10 questions to complete.

←

List

→

Return

Shaded items are complete.

1

2

3

4

5

6

7

8

9

10

End

Return

You have completed

questions

question

Your score is

Correct

Wrong

Partial-Credit

You have not finished your quiz. If you leave this page, your progress will be lost.